新高考數(shù)學(xué)一輪復(fù)習(xí) 專(zhuān)項(xiàng)分層精練第08課 指數(shù)函數(shù)（解析版）

第08課指數(shù)函數(shù)（分層專(zhuān)項(xiàng)精練）【一層練基礎(chǔ)】一、單選題1．（2023·全國(guó)·高三專(zhuān)題練習(xí)）英國(guó)著名數(shù)學(xué)家布魯克-泰勒以微積分學(xué)中將函數(shù)展開(kāi)成無(wú)窮級(jí)數(shù)的定理著稱(chēng)于世.在數(shù)學(xué)中，泰勒級(jí)數(shù)用無(wú)限連加式來(lái)表示一個(gè)函數(shù)，泰勒提出了適用于所有函數(shù)的泰勒級(jí)數(shù)，并建立了如下指數(shù)函數(shù)公式：SKIPIF1<0，其中SKIPIF1<0，則SKIPIF1<0的近似值為（精確到SKIPIF1<0）（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．（2023春·福建莆田·高二?？计谥校┰O(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<03．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知SKIPIF1<0則SKIPIF1<0等于（

）A．SKIPIF1<0 B．SKIPIF1<0 C．1 D．24．（2022秋·山東臨沂·高一?？茧A段練習(xí)）已知函數(shù)SKIPIF1<0（其中SKIPIF1<0）的圖象如圖所示，則函數(shù)SKIPIF1<0的圖象大致是（

）A．B．C．D．5．（2020·高一課時(shí)練習(xí)）SKIPIF1<0且SKIPIF1<0）是增函數(shù)，那么函數(shù)SKIPIF1<0的圖象大致是（

）A． B． C． D．6．（2020秋·江西吉安·高一校聯(lián)考期中）函數(shù)SKIPIF1<0，且SKIPIF1<0的圖象過(guò)一個(gè)定點(diǎn)，則這個(gè)定點(diǎn)坐標(biāo)是SKIPIF1<0SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<07．（2022秋·天津北辰·高三?？茧A段練習(xí)）函數(shù)SKIPIF1<0的部分圖象可能是（

）A．B． C． D．8．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知SKIPIF1<0且SKIPIF1<0，則函數(shù)SKIPIF1<0為奇函數(shù)的一個(gè)充分不必要條件是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<09．（2023·河北·高三學(xué)業(yè)考試）若SKIPIF1<0滿足不等式SKIPIF1<0，則函數(shù)SKIPIF1<0的值域是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<010．（2023·全國(guó)·高三專(zhuān)題練習(xí)）若函數(shù)f(x)＝SKIPIF1<0是R上的增函數(shù)，則實(shí)數(shù)a的取值范圍為（

）A．(1，＋∞) B．(1,8) C．(4,8) D．[4,8)11．（2023·江蘇南京·南京市第九中學(xué)?？寄M預(yù)測(cè)）已知SKIPIF1<0，c＝sin1，則a，b，c的大小關(guān)系是（

）A．c＜b＜a B．c＜a＜b C．a(chǎn)＜b＜c D．a(chǎn)＜c＜b12．（2022秋·廣東肇慶·高三肇慶市第一中學(xué)?？茧A段練習(xí)）若關(guān)于SKIPIF1<0的不等式SKIPIF1<0（SKIPIF1<0）恒成立，則實(shí)數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<013．（2023·全國(guó)·高三專(zhuān)題練習(xí)）要測(cè)定古物的年代，可以用放射性碳法：在動(dòng)植物的體內(nèi)都含有微量的放射性SKIPIF1<0．動(dòng)植物死亡后，停止了新陳代謝，SKIPIF1<0不再產(chǎn)生，且原來(lái)的SKIPIF1<0會(huì)自動(dòng)衰變．經(jīng)過(guò)5730年，它的殘余量只有原始量的一半．現(xiàn)用放射性碳法測(cè)得某古物中SKIPIF1<0含量占原來(lái)的SKIPIF1<0，推算該古物約是SKIPIF1<0年前的遺物（參考數(shù)據(jù)：SKIPIF1<0），則實(shí)數(shù)SKIPIF1<0的值為（

）A．12302 B．13304 C．23004 D．24034二、多選題14．（2024秋·湖北黃岡·高三浠水縣第一中學(xué)校考階段練習(xí)）已知函數(shù)SKIPIF1<0，則下列說(shuō)法正確的是（

）A．SKIPIF1<0為奇函數(shù) B．SKIPIF1<0為減函數(shù)C．SKIPIF1<0有且只有一個(gè)零點(diǎn) D．SKIPIF1<0的值域?yàn)镾KIPIF1<015．（2023·浙江紹興·統(tǒng)考模擬預(yù)測(cè)）預(yù)測(cè)人口的變化趨勢(shì)有多種方法，“直接推算法”使用的公式是SKIPIF1<0，其中SKIPIF1<0為預(yù)測(cè)期人口數(shù)，SKIPIF1<0為初期人口數(shù)，SKIPIF1<0為預(yù)測(cè)期內(nèi)人口年增長(zhǎng)率，SKIPIF1<0為預(yù)測(cè)期間隔年數(shù)，則（

）A．當(dāng)SKIPIF1<0，則這期間人口數(shù)呈下降趨勢(shì)B．當(dāng)SKIPIF1<0，則這期間人口數(shù)呈擺動(dòng)變化C．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的最小值為3D．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的最小值為3三、填空題16．（2022·全國(guó)·高三專(zhuān)題練習(xí)）已知函數(shù)SKIPIF1<0的值域?yàn)镽，則實(shí)數(shù)a的取值范圍是.17．（2022·全國(guó)·高三專(zhuān)題練習(xí)）下列函數(shù)是奇函數(shù)，且在SKIPIF1<0上單調(diào)遞增的是.①SKIPIF1<0②SKIPIF1<0③SKIPIF1<0④SKIPIF1<018．（2023春·北京石景山·高二統(tǒng)考期末）設(shè)函數(shù)SKIPIF1<0，則使得SKIPIF1<0成立的SKIPIF1<0的取值范圍是.19．（2014·甘肅天水·統(tǒng)考一模）下列5個(gè)判斷：①若SKIPIF1<0在SKIPIF1<0上增函數(shù)，則SKIPIF1<0；②函數(shù)SKIPIF1<0只有兩個(gè)零點(diǎn)；③函數(shù)SKIPIF1<0的值域是SKIPIF1<0；④函數(shù)SKIPIF1<0的最小值是1；⑤在同一坐標(biāo)系中函數(shù)SKIPIF1<0與SKIPIF1<0的圖像關(guān)于SKIPIF1<0軸對(duì)稱(chēng)．其中正確命題的序號(hào)【二層練綜合】一、單選題1．（2023·全國(guó)·高三專(zhuān)題練習(xí)）蘇格蘭數(shù)學(xué)家科林麥克勞林（ColinMaclaurin）研究出了著名的Maclaurin級(jí)數(shù)展開(kāi)式，受到了世界上頂尖數(shù)學(xué)家的廣泛認(rèn)可，下面是麥克勞林建立的其中一個(gè)公式：SKIPIF1<0，試根據(jù)此公式估計(jì)下面代數(shù)式SKIPIF1<0的近似值為（

）（可能用到數(shù)值SKIPIF1<0）A．2.322 B．4.785 C．4.755 D．1.0052．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知函數(shù)SKIPIF1<0是SKIPIF1<0上的偶函數(shù)，且SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0的值為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．23．（2020·安徽安慶·安慶市第七中學(xué)?？寄M預(yù)測(cè)）已知函數(shù)SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<04．（2022秋·寧夏銀川·高三銀川一中?？茧A段練習(xí)）若函數(shù)SKIPIF1<0是定義在SKIPIF1<0上的周期為2的奇函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0（

）A．0 B．2 C．4 D．-25．（2023·江蘇鎮(zhèn)江·揚(yáng)中市第二高級(jí)中學(xué)?？寄M預(yù)測(cè)）已知SKIPIF1<0是定義在SKIPIF1<0上的奇函數(shù)且滿足SKIPIF1<0為偶函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0（SKIPIF1<0且SKIPIF1<0）.若SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<06．（2023·全國(guó)·高三專(zhuān)題練習(xí)）函數(shù)SKIPIF1<0的圖像大致為（

）A． B．C． D．7．（2021秋·高一課時(shí)練習(xí)）已知正實(shí)數(shù)SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0，則A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<08．（2022秋·遼寧葫蘆島·高三校聯(lián)考期中）函數(shù)SKIPIF1<0（SKIPIF1<0，且SKIPIF1<0）的圖象恒過(guò)定點(diǎn)SKIPIF1<0，若點(diǎn)SKIPIF1<0在橢圓SKIPIF1<0（SKIPIF1<0，SKIPIF1<0）上，則SKIPIF1<0的最小值為（

）A．12 B．14 C．16 D．189．（2007·天津·高考真題）設(shè)SKIPIF1<0均為正數(shù)，且SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．則（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<010．（2023秋·山西朔州·高三懷仁市第一中學(xué)校?？计谀┮阎瘮?shù)SKIPIF1<0，則“SKIPIF1<0”是“函數(shù)SKIPIF1<0為偶函數(shù)”的（

）A．充分不必要條件 B．必要不充分條件C．充要條件 D．既不充分也不必要條件11．（2023·陜西西安·統(tǒng)考三模）如圖是下列四個(gè)函數(shù)中的某個(gè)函數(shù)在區(qū)間SKIPIF1<0上的大致圖象，則該函數(shù)是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<012．（2023春·高二課時(shí)練習(xí)）某微生物科研團(tuán)隊(duì)為了研究某種細(xì)菌的繁殖情況，工作人員配制了一種適合該細(xì)菌繁殖的營(yíng)養(yǎng)基質(zhì)用以培養(yǎng)該細(xì)菌，通過(guò)相關(guān)設(shè)備以及分析計(jì)算后得到：該細(xì)菌在前3個(gè)小時(shí)的細(xì)菌數(shù)SKIPIF1<0與時(shí)間SKIPIF1<0（單位：小時(shí)，且SKIPIF1<0）滿足回歸方程SKIPIF1<0（其中SKIPIF1<0為常數(shù)），若SKIPIF1<0，且前3個(gè)小時(shí)SKIPIF1<0與SKIPIF1<0的部分?jǐn)?shù)據(jù)如下表：SKIPIF1<0123SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<03個(gè)小時(shí)后，向該營(yíng)養(yǎng)基質(zhì)中加入某種細(xì)菌抑制劑，分析計(jì)算后得到細(xì)菌數(shù)SKIPIF1<0與時(shí)間SKIPIF1<0（單位：小時(shí)，且SKIPIF1<0）滿足關(guān)系式：SKIPIF1<0，在SKIPIF1<0時(shí)刻，該細(xì)菌數(shù)達(dá)到最大，隨后細(xì)菌個(gè)數(shù)逐漸減少，則SKIPIF1<0的值為（

）A．4 B．SKIPIF1<0 C．5 D．SKIPIF1<013．（2023春·山西大同·高二?？计谥校╃婌`大道是連接新余北站和新余城區(qū)的主干道，是新余對(duì)外交流的門(mén)戶之一，而仰天崗大橋就是這一條主干道的起點(diǎn)，其橋拱曲線形似懸鏈線，橋型優(yōu)美，被廣大市民們美稱(chēng)為“彩虹橋”，是我市的標(biāo)志性建筑之一，函數(shù)解析式為SKIPIF1<0，則下列關(guān)于SKIPIF1<0的說(shuō)法正確的是（

）A．SKIPIF1<0，SKIPIF1<0為奇函數(shù)B．SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增C．SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增D．SKIPIF1<0，SKIPIF1<0有最小值114．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知SKIPIF1<0為偶函數(shù)，SKIPIF1<0為奇函數(shù)，且滿足SKIPIF1<0.若存在SKIPIF1<0，使得不等式SKIPIF1<0有解，則實(shí)數(shù)SKIPIF1<0的最大值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．1 D．-115．（2021·陜西西安·統(tǒng)考三模）射線測(cè)厚技術(shù)原理公式為SKIPIF1<0，其中SKIPIF1<0分別為射線穿過(guò)被測(cè)物前后的強(qiáng)度，SKIPIF1<0是自然對(duì)數(shù)的底數(shù)，SKIPIF1<0為被測(cè)物厚度，SKIPIF1<0為被測(cè)物的密度，SKIPIF1<0是被測(cè)物對(duì)射線的吸收系數(shù).工業(yè)上通常用镅241（SKIPIF1<0）低能SKIPIF1<0射線測(cè)量鋼板的厚度.若這種射線對(duì)鋼板的半價(jià)層厚度為0.8，鋼的密度為7.6，則這種射線的吸收系數(shù)為（

）（注：半價(jià)層厚度是指將已知射線強(qiáng)度減弱為一半的某種物質(zhì)厚度，SKIPIF1<0，結(jié)果精確到0.001）A．0.110 B．0.112 C．SKIPIF1<0 D．SKIPIF1<0二、多選題16．（2022·全國(guó)·高一期末）若兩函數(shù)的定義域、單調(diào)區(qū)間、奇偶性、值域都相同，則稱(chēng)這兩函數(shù)為“伙伴函數(shù)”.下列函數(shù)中與函數(shù)SKIPIF1<0不是“伙伴函數(shù)”是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<017．（2022秋·廣東廣州·高一廣州市真光中學(xué)校考期中）若實(shí)數(shù)SKIPIF1<0，SKIPIF1<0滿足SKIPIF1<0則下列關(guān)系式中可能成立的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<018．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知實(shí)數(shù)SKIPIF1<0滿足SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<019．（2022秋·山西太原·高一?？茧A段練習(xí)）已知SKIPIF1<0，且SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0三、填空題20．（2023·全國(guó)·高三專(zhuān)題練習(xí)）函數(shù)SKIPIF1<0在SKIPIF1<0的值域?yàn)椋?1．（2020·高一課時(shí)練習(xí)）定義區(qū)間[x1，x2]的長(zhǎng)度為x2－x1，已知函數(shù)f(x)＝3|x|的定義域?yàn)閇a，b]，值域?yàn)閇1，9]，則區(qū)間[a，b]長(zhǎng)度的最小值為.22．（2022秋·四川成都·高三樹(shù)德中學(xué)校考階段練習(xí)）已知函數(shù)SKIPIF1<0，則不等式SKIPIF1<0的解集是.23．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知實(shí)數(shù)SKIPIF1<0，SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0．【三層練能力】一、單選題1．（2022·天津北辰·?？寄M預(yù)測(cè)）已知SKIPIF1<0且SKIPIF1<0，函數(shù)SKIPIF1<0在SKIPIF1<0上是單調(diào)函數(shù)，若關(guān)于SKIPIF1<0的方程SKIPIF1<0恰有2個(gè)互異的實(shí)數(shù)解，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<02．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知函數(shù)SKIPIF1<0，存在實(shí)數(shù)SKIPIF1<0使得SKIPIF1<0成立，若正整數(shù)SKIPIF1<0的最大值為6，則SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0二、多選題3．（2023秋·黑龍江大慶·高三鐵人中學(xué)校考期末）當(dāng)SKIPIF1<0時(shí)，不等式SKIPIF1<0成立．若SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<04．（2023春·北京海淀·高二北京交通大學(xué)附屬中學(xué)校考期中）設(shè)函數(shù)SKIPIF1<0，則下列選項(xiàng)正確的是（

）A．SKIPIF1<0為奇函數(shù)B．SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)C．SKIPIF1<0的最小值為SKIPIF1<0D．若SKIPIF1<0有兩個(gè)不等實(shí)根，則SKIPIF1<0，且SKIPIF1<0三、填空題5．（2023·四川南充·閬中中學(xué)?？级＃┮阎猄KIPIF1<0，若存在SKIPIF1<0，使得SKIPIF1<0，則SKIPIF1<0的取值范圍為.6．（2022春·上海浦東新·高三上海市建平中學(xué)?？茧A段練習(xí)）將SKIPIF1<0的圖象向右平移2個(gè)單位后得曲線SKIPIF1<0，將函數(shù)SKIPIF1<0的圖象向下平移2個(gè)單位后得曲線SKIPIF1<0，SKIPIF1<0與SKIPIF1<0關(guān)于SKIPIF1<0軸對(duì)稱(chēng)．若SKIPIF1<0的最小值為SKIPIF1<0且SKIPIF1<0，則實(shí)數(shù)SKIPIF1<0的取值范圍為．【一層練基礎(chǔ)】參考答案1．C【分析】應(yīng)用題設(shè)泰勒展開(kāi)式可得SKIPIF1<0,隨著SKIPIF1<0的增大，數(shù)列SKIPIF1<0遞減且靠后各項(xiàng)無(wú)限接近于SKIPIF1<0，即可估計(jì)SKIPIF1<0的近似值.【詳解】計(jì)算前四項(xiàng)，在千分位上四舍五入由題意知:SKIPIF1<0SKIPIF1<0故選：C2．D【詳解】SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，根據(jù)SKIPIF1<0在SKIPIF1<0上是增函數(shù)，所以SKIPIF1<0，即SKIPIF1<0.故選：D.3．C【分析】根據(jù)分段函數(shù)的解析式，結(jié)合對(duì)應(yīng)區(qū)間求SKIPIF1<0即可.【詳解】∵26>4，∴SKIPIF1<0，又SKIPIF1<0，∴SKIPIF1<0.故選：C.4．C【分析】根據(jù)二次函數(shù)圖象特點(diǎn)，結(jié)合指數(shù)型函數(shù)圖象的特點(diǎn)進(jìn)行判斷即可.【詳解】SKIPIF1<0的函數(shù)圖象與SKIPIF1<0軸的交點(diǎn)的橫坐標(biāo)為SKIPIF1<0的兩個(gè)根，由SKIPIF1<0可得兩根為a，b，觀察SKIPIF1<0的圖象，可得其與軸的兩個(gè)交點(diǎn)分別在區(qū)間SKIPIF1<0與SKIPIF1<0上，又∵SKIPIF1<0，∴SKIPIF1<0，SKIPIF1<0，SKIPIF1<0由可知，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0為增函數(shù)，又由SKIPIF1<0得SKIPIF1<0的圖象與y軸的交點(diǎn)在x軸上方，分析選項(xiàng)可得C符合這兩點(diǎn)．故選：C．5．D【分析】先根據(jù)函數(shù)SKIPIF1<0且SKIPIF1<0）的單調(diào)性判斷底數(shù)SKIPIF1<0的范圍，得到函數(shù)SKIPIF1<0的圖象，再利用圖象平移得到函數(shù)SKIPIF1<0的圖象．【詳解】解；∵SKIPIF1<0可變形為SKIPIF1<0，若它是增函數(shù)，則SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0為過(guò)點(diǎn)（1，0）的減函數(shù)，∴SKIPIF1<0為過(guò)點(diǎn)（1，0）的增函數(shù)，∵SKIPIF1<0圖象為SKIPIF1<0圖象向左平移1個(gè)單位長(zhǎng)度，∴SKIPIF1<0圖象為過(guò)（0，0）點(diǎn)的增函數(shù)，故選D．【點(diǎn)睛】本題考查了指對(duì)數(shù)函數(shù)的單調(diào)性，以及圖象的平移變化，做題時(shí)要認(rèn)真觀察．6．B【詳解】試題分析：令SKIPIF1<0得SKIPIF1<0SKIPIF1<0時(shí)SKIPIF1<0，所以過(guò)定點(diǎn)SKIPIF1<0考點(diǎn)：指數(shù)函數(shù)性質(zhì)7．C【分析】根據(jù)函數(shù)的奇偶性排除AB，再根據(jù)SKIPIF1<0趨近于SKIPIF1<0時(shí)的值判斷即可【詳解】因?yàn)镾KIPIF1<0，故SKIPIF1<0為奇函數(shù)，排除AB，又當(dāng)SKIPIF1<0趨近于SKIPIF1<0時(shí)，SKIPIF1<0遠(yuǎn)遠(yuǎn)大于SKIPIF1<0，所有函數(shù)逐漸趨近于0，排除D故選：C8．C【分析】先利用奇函數(shù)的性質(zhì)SKIPIF1<0得到SKIPIF1<0，然后在利用定義驗(yàn)證此時(shí)函數(shù)SKIPIF1<0為奇函數(shù)，從而得到函數(shù)SKIPIF1<0為奇函數(shù)的充分必要條件是SKIPIF1<0，進(jìn)而根據(jù)充分不必要條件的概念作出判定.【詳解】若函數(shù)SKIPIF1<0為奇函數(shù)，由于函數(shù)SKIPIF1<0的定義域?yàn)镽，∴SKIPIF1<0，∴SKIPIF1<0,即SKIPIF1<0,∴SKIPIF1<0∴SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0為奇函數(shù)的充分必要條件是SKIPIF1<0或SKIPIF1<0，SKIPIF1<0是SKIPIF1<0的非充分非必要條件；SKIPIF1<0是SKIPIF1<0的非充分非必要條件；SKIPIF1<0是SKIPIF1<0的充分不必要條件；故選：C.9．B【分析】先將不等式左右兩邊化為底數(shù)相同，再由指數(shù)函數(shù)的單調(diào)性解不等式即可求得SKIPIF1<0的范圍，再由指數(shù)函數(shù)的單調(diào)性即可求值域.【詳解】由SKIPIF1<0可得SKIPIF1<0，因?yàn)镾KIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0即SKIPIF1<0，解得：SKIPIF1<0，所以SKIPIF1<0，即函數(shù)SKIPIF1<0的值域是SKIPIF1<0，故選：B.10．D【分析】根據(jù)函數(shù)的單調(diào)性給出不等式組，求解參數(shù)的取值范圍即可.【詳解】由題意得SKIPIF1<0SKIPIF1<0解得4≤a＜8.故選：D.11．D【分析】由對(duì)數(shù)的運(yùn)算法則求出a,然后根據(jù)指數(shù)函數(shù)與正弦函數(shù)的單調(diào)性分別對(duì)b,c進(jìn)行放縮，最后求得答案.【詳解】由題意，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0.故選：D.12．B【分析】根據(jù)指數(shù)函數(shù)的性質(zhì)，參變分離可得SKIPIF1<0恒成立，再根據(jù)冪函數(shù)的性質(zhì)計(jì)算可得；【詳解】解：因?yàn)镾KIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0恒成立，即SKIPIF1<0恒成立，因?yàn)镾KIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0；故選：B13．B【分析】設(shè)SKIPIF1<0每年的衰變率為SKIPIF1<0，古物中原SKIPIF1<0的含量為SKIPIF1<0，然后根據(jù)半衰期，建立方程，將已知條件帶入取對(duì)數(shù)，利用對(duì)數(shù)性質(zhì)運(yùn)算即可.【詳解】設(shè)SKIPIF1<0每年的衰變率為SKIPIF1<0，古物中原SKIPIF1<0的含量為SKIPIF1<0，由半衰期，得SKIPIF1<0．所以SKIPIF1<0，即SKIPIF1<0．由題意，知SKIPIF1<0，即SKIPIF1<0．于是SKIPIF1<0．所以SKIPIF1<0．故選：B．14．AC【分析】化簡(jiǎn)函數(shù)解析式，分析函數(shù)的奇偶性，單調(diào)性，值域，零點(diǎn)即可求解.【詳解】SKIPIF1<0,SKIPIF1<0,SKIPIF1<0SKIPIF1<0,故SKIPIF1<0為奇函數(shù)，又SKIPIF1<0，SKIPIF1<0在R上單調(diào)遞增，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，即函數(shù)值域?yàn)镾KIPIF1<0令SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，故函數(shù)有且只有一個(gè)零點(diǎn)0.綜上可知，AC正確，BD錯(cuò)誤.故選：AC15．AC【分析】由指數(shù)函數(shù)的性質(zhì)確定函數(shù)的增減性可判斷A，B；分別代入SKIPIF1<0和SKIPIF1<0，解指數(shù)不等式可判斷C，D.【詳解】SKIPIF1<0，由指數(shù)函數(shù)的性質(zhì)可知：SKIPIF1<0是關(guān)于n的單調(diào)遞減函數(shù)，即人口數(shù)呈下降趨勢(shì)，故A正確，B不正確；SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0的最小值為3，故C正確；SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0的最小值為2，故D不正確；故選：AC.16．SKIPIF1<0【分析】首先分別求分段函數(shù)兩段的值域，再根據(jù)值域?yàn)镾KIPIF1<0，列式求實(shí)數(shù)a的取值范圍.【詳解】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，因?yàn)楹瘮?shù)的值域?yàn)镾KIPIF1<0，所以SKIPIF1<0，解得：SKIPIF1<0.故答案為：SKIPIF1<017．①③【分析】根據(jù)奇函數(shù)的定義及基本初等函數(shù)的單調(diào)性，對(duì)選項(xiàng)進(jìn)行一一判斷，即可得答案；【詳解】①SKIPIF1<0，所以函數(shù)SKIPIF1<0是奇函數(shù)，又SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故正確；②SKIPIF1<0，所以SKIPIF1<0，所以函數(shù)SKIPIF1<0不是奇函數(shù)，故錯(cuò)誤；③SKIPIF1<0，所以函數(shù)SKIPIF1<0是奇函數(shù)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故正確；④SKIPIF1<0SKIPIF1<0，所以函數(shù)SKIPIF1<0是偶函數(shù)，故錯(cuò)誤；故答案為：①③18．SKIPIF1<0【分析】分SKIPIF1<0和SKIPIF1<0兩種情況討論從而解不等式SKIPIF1<0即可.【詳解】當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0，又因?yàn)镾KIPIF1<0，所以SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0，又因?yàn)镾KIPIF1<0，所以SKIPIF1<0.所以滿足SKIPIF1<0成立的SKIPIF1<0的取值范圍為SKIPIF1<0.故答案為：SKIPIF1<0.19．④⑤【分析】根據(jù)函數(shù)性質(zhì)逐個(gè)分析求解即可.【詳解】①若SKIPIF1<0在SKIPIF1<0上增函數(shù)，所以只需SKIPIF1<0，即SKIPIF1<0，故①不正確；②畫(huà)出SKIPIF1<0和SKIPIF1<0負(fù)半軸的圖象，根據(jù)圖象可觀察在負(fù)半軸有一個(gè)零點(diǎn)，又SKIPIF1<0，SKIPIF1<0，所以函數(shù)SKIPIF1<0只有三個(gè)零點(diǎn)，故②不正確；③因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以值域是SKIPIF1<0，故③不正確；④因?yàn)镾KIPIF1<0，所以SKIPIF1<0，即最小值是SKIPIF1<0，故④正確；⑤因?yàn)閮蓚€(gè)函數(shù)定義域相同，且用SKIPIF1<0代替SKIPIF1<0中的SKIPIF1<0，兩個(gè)解析式完全相同，所以圖象關(guān)于SKIPIF1<0軸對(duì)稱(chēng)，故⑤正確.故答案為：④⑤.【二層練綜合】參考答案1．C【分析】由題目觀察可知SKIPIF1<0，代入SKIPIF1<0即可發(fā)現(xiàn)解法.【詳解】SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0所以SKIPIF1<0=4.755故選：C2．C【分析】由函數(shù)SKIPIF1<0的圖像關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)得到SKIPIF1<0，結(jié)合SKIPIF1<0是偶函數(shù)得到SKIPIF1<0，進(jìn)一步得到SKIPIF1<0的周期是4，再利用周期性計(jì)算即可得到答案.【詳解】因?yàn)镾KIPIF1<0是SKIPIF1<0上的偶函數(shù)，所以SKIPIF1<0，又SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)，則SKIPIF1<0，所以SKIPIF1<0，則SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0是周期函數(shù)，且周期SKIPIF1<0，由SKIPIF1<0時(shí)，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0.故選：C3．D【詳解】分析：先化簡(jiǎn)SKIPIF1<0得到SKIPIF1<0，再求SKIPIF1<0的值.詳解：由題得SKIPIF1<0SKIPIF1<0所以SKIPIF1<0故答案為D點(diǎn)睛：（1）本題主要考查函數(shù)求值和指數(shù)對(duì)數(shù)運(yùn)算，意在考查學(xué)生對(duì)這些基礎(chǔ)知識(shí)的掌握能力和運(yùn)算能力.（2）解答本題的關(guān)鍵是整體代入求值.4．D【分析】由已知，利用奇函數(shù)及周期性求SKIPIF1<0，SKIPIF1<0的函數(shù)值，即可求目標(biāo)式的值.【詳解】∵SKIPIF1<0是定義在SKIPIF1<0上的奇函數(shù)，∴SKIPIF1<0，又SKIPIF1<0在SKIPIF1<0上的周期為2，∴SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0.故選：D.5．B【分析】根據(jù)已知條件可得SKIPIF1<0的對(duì)稱(chēng)中心SKIPIF1<0，對(duì)稱(chēng)軸SKIPIF1<0，可得SKIPIF1<0為SKIPIF1<0的一個(gè)周期，由SKIPIF1<0、SKIPIF1<0以及SKIPIF1<0列關(guān)于SKIPIF1<0的方程組，進(jìn)而可得SKIPIF1<0時(shí)，SKIPIF1<0的解析式，再利用周期性即可求解.【詳解】因?yàn)镾KIPIF1<0為奇函數(shù)，所以SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0中心對(duì)稱(chēng)，因?yàn)镾KIPIF1<0為偶函數(shù)，所以SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱(chēng).根據(jù)條件可知SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0為SKIPIF1<0的一個(gè)周期，則SKIPIF1<0，又因?yàn)镾KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0（舍），所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0，故選：B.6．A【分析】根據(jù)奇偶性和值域，運(yùn)用排除法求解.【詳解】設(shè)SKIPIF1<0，則有SKIPIF1<0，SKIPIF1<0是奇函數(shù)，排除D；SKIPIF1<0，排除B；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，排除C；故選：A.7．B【分析】在同一坐標(biāo)系內(nèi)，分別作出函數(shù)SKIPIF1<0的圖象，結(jié)合圖象，即可求解．【詳解】由題意，在同一坐標(biāo)系內(nèi)，分別作出函數(shù)SKIPIF1<0的圖象，結(jié)合圖象可得：SKIPIF1<0，故選B．【點(diǎn)睛】本題主要考查了指數(shù)函數(shù)、對(duì)數(shù)函數(shù)的圖象與性質(zhì)的應(yīng)用，其中解中熟記指數(shù)函數(shù)、對(duì)數(shù)函數(shù)的圖象，結(jié)合圖象求解是解答的關(guān)鍵，著重考查了數(shù)形結(jié)合思想，以及推理與運(yùn)算能力，屬于中檔試題．8．C【分析】求出SKIPIF1<0的坐標(biāo)代入橢圓方程，再將SKIPIF1<0化為積為定值的形式，利用基本不等式可求得結(jié)果.【詳解】由SKIPIF1<0，即SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0，因?yàn)辄c(diǎn)SKIPIF1<0在橢圓SKIPIF1<0上，所以SKIPIF1<0（SKIPIF1<0，SKIPIF1<0），所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號(hào)成立.故選：C【點(diǎn)睛】易錯(cuò)點(diǎn)睛：利用基本不等式求最值時(shí)，要注意其必須滿足的三個(gè)條件：（1）“一正二定三相等”“一正”就是各項(xiàng)必須為正數(shù)；（2）“二定”就是要求和的最小值，必須把構(gòu)成和的二項(xiàng)之積轉(zhuǎn)化成定值；要求積的最大值，則必須把構(gòu)成積的因式的和轉(zhuǎn)化成定值；（3）“三相等”是利用基本不等式求最值時(shí)，必須驗(yàn)證等號(hào)成立的條件，若不能取等號(hào)則這個(gè)定值就不是所求的最值，這也是最容易發(fā)生錯(cuò)誤的地方.9．A【分析】試題分析：在同一坐標(biāo)系中分別畫(huà)出SKIPIF1<0SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的圖象，SKIPIF1<0與SKIPIF1<0的交點(diǎn)的橫坐標(biāo)為SKIPIF1<0，SKIPIF1<0與SKIPIF1<0的圖象的交點(diǎn)的橫坐標(biāo)為SKIPIF1<0，SKIPIF1<0與SKIPIF1<0的圖象的交點(diǎn)的橫坐標(biāo)為SKIPIF1<0，從圖象可以看出．考點(diǎn)：指數(shù)函數(shù)、對(duì)數(shù)函數(shù)圖象和性質(zhì)的應(yīng)用．【方法點(diǎn)睛】一般一個(gè)方程中含有兩個(gè)以上的函數(shù)類(lèi)型，就要考慮用數(shù)形結(jié)合求解，在同一坐標(biāo)系中畫(huà)出兩函數(shù)圖象的交點(diǎn)，函數(shù)圖象的交點(diǎn)的橫坐標(biāo)即為方程的解．【詳解】10．A【分析】由給定函數(shù)SKIPIF1<0，求出SKIPIF1<0為偶函數(shù)時(shí)的a值，再利用充分條件、必要條件的定義判斷作答.【詳解】函數(shù)SKIPIF1<0定義域?yàn)镽，函數(shù)SKIPIF1<0為偶函數(shù)，則SKIPIF1<0,SKIPIF1<0，而SKIPIF1<0不恒為0，因此，SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，所以“SKIPIF1<0”是“函數(shù)SKIPIF1<0為偶函數(shù)”的充分不必要條件.故選：A11．A【分析】根據(jù)給定的函數(shù)圖象特征，利用函數(shù)的奇偶性排除BC；利用SKIPIF1<0的正負(fù)即可判斷作答.【詳解】對(duì)于B，SKIPIF1<0，SKIPIF1<0，函數(shù)SKIPIF1<0是偶函數(shù)，B不是；對(duì)于C，SKIPIF1<0，SKIPIF1<0，函數(shù)SKIPIF1<0是偶函數(shù)，C不是；對(duì)于D，SKIPIF1<0，SKIPIF1<0，D不是；對(duì)于A，SKIPIF1<0，SKIPIF1<0，函數(shù)SKIPIF1<0是奇函數(shù)，且SKIPIF1<0，A符合題意.故選：A12．A【分析】根據(jù)給定條件，求出樣本中心點(diǎn)求出b值，再分段討論y的最大值情況作答.【詳解】依題意，SKIPIF1<0，SKIPIF1<0，由SKIPIF1<0，SKIPIF1<0，得SKIPIF1<0，且SKIPIF1<0經(jīng)過(guò)點(diǎn)SKIPIF1<0，于是得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增，則當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，令SKIPIF1<0，SKIPIF1<0，求導(dǎo)得：SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0上單調(diào)遞減，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，而SKIPIF1<0，因此當(dāng)SKIPIF1<0時(shí)，細(xì)菌數(shù)SKIPIF1<0取最大值，所以SKIPIF1<0的值為4.故選：A13．B【分析】根據(jù)函數(shù)奇偶性的定義及復(fù)合函數(shù)的單調(diào)性逐一判定即可.【詳解】由題意易得SKIPIF1<0定義域?yàn)镽，SKIPIF1<0，即SKIPIF1<0為偶函數(shù)，故A錯(cuò)誤；令SKIPIF1<0，則SKIPIF1<0且SKIPIF1<0隨SKIPIF1<0增大而增大，此時(shí)SKIPIF1<0，由對(duì)勾函數(shù)的單調(diào)性得SKIPIF1<0單調(diào)遞增，根據(jù)復(fù)合函數(shù)的單調(diào)性原則得SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故B正確；結(jié)合A項(xiàng)得SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，故C錯(cuò)誤；結(jié)合B項(xiàng)及對(duì)勾函數(shù)的性質(zhì)得SKIPIF1<0，故D錯(cuò)誤.故選：B.14．A【解析】由題意得出SKIPIF1<0、SKIPIF1<0的解析式，不等式恒成立，采用分離參數(shù)法，可得SKIPIF1<0轉(zhuǎn)化為求函數(shù)的最值，求出函數(shù)SKIPIF1<0的最大值即可.【詳解】SKIPIF1<0SKIPIF1<0為偶函數(shù)，SKIPIF1<0為奇函數(shù)，且SKIPIF1<0①SKIPIF1<0②①②兩式聯(lián)立可得SKIPIF1<0，SKIPIF1<0.由SKIPIF1<0得SKIPIF1<0，∵SKIPIF1<0在SKIPIF1<0為增函數(shù)，∴SKIPIF1<0，故選：A.【點(diǎn)睛】本題主要考查函數(shù)奇偶性的應(yīng)用、考查了不等式存在有解問(wèn)題以及函數(shù)的單調(diào)性求最值，注意分離參數(shù)法的應(yīng)用，此題屬于中檔題.15．C【解析】根據(jù)題意知,SKIPIF1<0，代入公式SKIPIF1<0,求出SKIPIF1<0即可.【詳解】由題意可得,SKIPIF1<0因?yàn)镾KIPIF1<0,所以SKIPIF1<0,即SKIPIF1<0.所以這種射線的吸收系數(shù)為SKIPIF1<0.故選:C【點(diǎn)睛】本題主要考查知識(shí)的遷移能力,把數(shù)學(xué)知識(shí)與物理知識(shí)相融合;重點(diǎn)考查指數(shù)型函數(shù),利用指數(shù)的相關(guān)性質(zhì)來(lái)研究指數(shù)型函數(shù)的性質(zhì),以及解指數(shù)型方程；屬于中檔題.16．BD【分析】分析各選項(xiàng)中的函數(shù)以及函數(shù)SKIPIF1<0的定義域、單調(diào)區(qū)間、奇偶性、值域都相同，結(jié)合題中定義可得出合適的選項(xiàng).【詳解】函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，單調(diào)遞增區(qū)間為SKIPIF1<0，遞減區(qū)間為SKIPIF1<0，該函數(shù)為偶函數(shù)，值域?yàn)镾KIPIF1<0.對(duì)于A選項(xiàng)，令SKIPIF1<0，該函數(shù)的定義域?yàn)镾KIPIF1<0，SKIPIF1<0，函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為SKIPIF1<0，遞減區(qū)間為SKIPIF1<0，因?yàn)镾KIPIF1<0，即函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0.SKIPIF1<0，即函數(shù)SKIPIF1<0為偶函數(shù)，A滿足條件；對(duì)于B選項(xiàng)，由SKIPIF1<0可得SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，故函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0，B不滿足條件；對(duì)于C選項(xiàng)，令SKIPIF1<0，該函數(shù)的定義域?yàn)镾KIPIF1<0，SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0且SKIPIF1<0不恒為零，所以，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)函數(shù)SKIPIF1<0單調(diào)遞減，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)函數(shù)SKIPIF1<0單調(diào)遞增，SKIPIF1<0，故函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0，因?yàn)镾KIPIF1<0，即函數(shù)SKIPIF1<0為偶函數(shù)，C滿足條件；對(duì)于D選項(xiàng)，函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，D不滿足條件.故選：BD.17．BCD【分析】先構(gòu)造函數(shù)SKIPIF1<0，SKIPIF1<0，判斷函數(shù)單調(diào)性并作圖，再判斷SKIPIF1<0，SKIPIF1<0，結(jié)合圖象即得兩個(gè)函數(shù)值相等時(shí)對(duì)應(yīng)自變量的大小關(guān)系.【詳解】設(shè)SKIPIF1<0，SKIPIF1<0，由初等函數(shù)的性質(zhì)，可得SKIPIF1<0，SKIPIF1<0都是單調(diào)遞增函數(shù)，畫(huà)出函數(shù)SKIPIF1<0，SKIPIF1<0的圖象，如圖所示，根據(jù)圖象可知，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，依題意不妨設(shè)SKIPIF1<0，則SKIPIF1<0分別是直線SKIPIF1<0與函數(shù)SKIPIF1<0圖象的交點(diǎn)的橫坐標(biāo).當(dāng)SKIPIF1<0時(shí)，若SKIPIF1<0，則SKIPIF1<0，故A不正確；當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，若SKIPIF1<0，則SKIPIF1<0或SKIPIF1<0，故B正確；當(dāng)SKIPIF1<0時(shí)，若SKIPIF1<0，則SKIPIF1<0，故C正確；當(dāng)SKIPIF1<0時(shí)，若SKIPIF1<0，則SKIPIF1<0，故D正確．故選：BCD．【點(diǎn)睛】本題的解題關(guān)鍵在于結(jié)合圖象判斷SKIPIF1<0，SKIPIF1<0，即可判斷兩個(gè)函數(shù)值相等時(shí)對(duì)應(yīng)自變量的大小關(guān)系，突破難點(diǎn).18．BCD【分析】A.由SKIPIF1<0得到SKIPIF1<0判斷；BC.由SKIPIF1<0，得到SKIPIF1<0判斷；D.由SKIPIF1<0，得到SKIPIF1<0，令SKIPIF1<0，用導(dǎo)數(shù)法判斷.【詳解】由SKIPIF1<0得SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，選項(xiàng)SKIPIF1<0錯(cuò)誤；因?yàn)镾KIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，選項(xiàng)SKIPIF1<0正確，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.令SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0